The Co-efficient of \(x^{3}\) in \(\left(1-x+x^{2}\right)^{5}\) is (a) \(-30\) (b) \(-20\) (c) \(-10\) (d) 30

We have the expression \((1-x+x^2)^5\). We can apply the binomial theorem to expand the expression which will give us various terms containing different power combinations of x. Using the binomial theorem, the general term for the expansion is given by: \(T_{r+1} = \binom{n}{r} (a)^{n-r} (b)^r\), where n = 5 and a = 1-x, b = x^2 We want to find the term having \(x^3\) in the expansion of \((1-x+x^2)^5\). So we need to determine the value of r when the power of x is 3.

For the required term having \(x^3\), we need the general term of the expansion to be of the form: \(k * x^3\) where k is the coefficient of the \(x^3\) term. Now we need to find the value of r that gives us the \(x^3\) term. We know, \(T_{r+1} = \binom{n}{r} (1-x)^{n-r} (x^2)^r\) So, using n = 5, we write the general term as: \(T_{r+1} = \binom{5}{r} (1-x)^{5-r} (x^2)^r\) We need to find the value of r such that the power of x in this term is 3. Let's first analyze the powers of x in different terms of the equation: 1. (1-x) has 1 power of x 2. (x^2) has 2 powers of x. Now, consider different possibilities of powers of x: - Case 1: If there is 1 power of x from (1-x) and 0 power of x from (x^2), then the power of x would be 1, which is not equal to 3. - Case 2: If there is 1 power of x from (1-x) and 1 power of x from (x^2), then the power of x would be 1+2=3, which is equal to 3. For this case, r=1. - Case 3: If there are 0 powers of x from (1-x) and 2 powers of x from (x^2), then the power of x would be 2, which is not equal to 3. We can see that only in case 2, we get the power of x equal to 3. So, we will use r=1 to get the term with \(x^3\).

Now we have determined r=1 to get the term with \(x^3\). So, let's plug in the value of r in the general term to find the \(x^3\) term: \(T_{1+1} = \binom{5}{1} (1-x)^{5-1} (x^2)^1\) \(T_2 = \binom{5}{1} (1-x)^4 (x^2)\) \(T_2 = 5 (1-4x+6x^2-4x^3+x^4) (x^2)\) Now, multiply the above equation to get the term with \(x^3\): \(T_2 = 5x^2 - 20x^3 + 30x^4 - 20x^5 + 5x^6\) We can see that the coefficient of \(x^3\) in the above equation is -20. So, the coefficient of \(x^3\) in \((1-x+x^2)^5\) is -20. The correct answer is (b) -20.